I need to create a regex expression for the following scenario.
It can have only numbers and only one dot or comma.
First part can have one to three digits.
The second part can be a dot or a comma.
The third part can have one to two digits.
The valid scenarios are
123,12
123.12
123,1
123
12,12
12.12
1,12
1.12
1,1
1.1
1
I came up so far with this expression
\d{1,3}(?:[.,]\d{1,2})?
but it doesn't work well. For example the input is 11:11 is marked as valid.
You need to put anchors around your expression:
^\d{1,3}(?:[.,]\d{1,2})?$
^ will match the start of the string
$ will match the end of the string
If those anchors are missing, it will partially match on your string, since the last part is optional, means on "11:11" it can match on the digits before the colon and a second match will be on the digits after the colon.
Try to use ^ and $:
^\d{1,3}(?:[.,]\d{1,2})?$
^ The match must start at the beginning of the string or line.
$ The match must occur at the end of the string or before \n at the end of the line or string.
Related
This seems like it should be easy, but I'm not so good with regex, and this doesn't seem to be easy to find on google.
I need a regex that starts with the string 'SP-multiple digits' and ends with the string '- multiple digits'
For example i have to match '-12' in "Sp-1234-12".
My attempt was: [^*-]*$ -> This case matches everything after the minus but i need the minus included.
For that digit and hyphen format, you could use a capture group for the part of the string that you want:
^Sp(?:-\d+)*(-\d+)$
Explanation
^ Start of string
Sp Match literally
(?:-\d+)* Optionally repeat - and 1+ digits
(-\d+) Capture group 1, match - and 1+ digits
$ End of string
Regex demo
Note that in C# you can use [0-9] instead of \d to match only digits 0-9
I am trying to write a regex to handle these cases
contains only alphanumeric with minimum of 2 alpha characters(numbers are optional).
only special character allowed is hyphen.
cannot be all same letter ignoring hyphen.
cannot be all hyphens
cannot be all numeric
My regex: (?=[^A-Za-z]*[A-Za-z]){2}^[\w-]{6,40}$
Above regex works for most of the scenarios except 1) & 3).
Can anyone suggest me to fix this. I am stuck in this.
Regards,
Sajesh
Rule 1 eliminates rule 4 and 5: It can neither contain only hyphens, nor only digits.
/^(?=[a-z\d-]{6,40}$)[\d-]*([a-z]).*?(?!\1)[a-z].*$/i
(?=[a-z\d-]{6,40}$) look ahead for specified characters from 6 to 40
([a-z]).*?(?!\1)[a-z] checks for two letters and at least one different
See this demo at regex101
This pattern with i flag considers A and a as the "same" letter (caseless matching) and will require another alpbhabet. For case sensitive matching here another demo at regex101.
You can use
^(?!\d+$)(?!-+$)(?=(?:[\d-]*[A-Za-z]){2})(?![\d-]*([A-Za-z])(?:[\d-]*\1)+[\d-]*$)[A-Za-z\d-]{6,40}$
See the regex demo. If you use it in C# or PHP, consider replacing ^ with \A and $ with \z to make sure you match the entire string even in case there is a trailing newline.
Details:
^ - start of string
(?!\d+$) - fail the match if the string only consists of digits
(?!-+$) - fail the match if the string only consists of hyphens
(?=(?:[\d-]*[A-Za-z]){2}) - there must be at least two ASCII letters after any zero or more digits or hyphens
(?![\d-]*([A-Za-z])(?:[\d-]*\1)+[\d-]*$) - fail the match if the string contains two or more identical letters (the + after (?:[\d-]*\1) means there can be any one letter)
[A-Za-z\d-]{6,40} - six to forty alphanumeric or hyphen chars
$ - end of string. (\z might be preferable.)
i have this expression and i need to make sure to include at least one non-alphabetic character
^(?!.*(.)\1)\S{8,12}$
testhis invalid
testhis7 valid
testhis# valid
You could use a positive lookahead asserting at least 1 char other than a-zA-Z
^(?!.*(.)\1)(?=.*[^\sa-zA-Z])\S{8,12}$
Explanation
^ Start of string
(?!.*(.)\1) Assert not 2 consecutive chars
(?=.*[^\sa-zA-Z]) Assert 1 char other than a whitespace char and a-zA-Z
\S{8,12} Match 8-12 non whitespace chars
$ End of string
Regex demo
Another option is to use \P{L} to assert any char other than any kind of letter from any language
^(?!.*(.)\1)(?=.*\P{L})\S{8,12}$
Regex demo
You can just check for the special character (as matched by [\p{P}\p{S}]) in positive lookahead (?=.*[\p{P}\p{S}]), which gives you the regex:
^(?!.*(.)\1)(?=.*[\p{P}\p{S}])\S{8,12}$
See online demo
You can also replace [\p{P}\p{S}] by [!"\#$%&'()*+,\-./:;<=>?#\[\\\]^_‘{|}~], or any other character set that list all the characters that you want to count as being "special characters".
It's better to do it with separate if-statements. This way you'll have exact information what is missing in the value. With regexps you'll only get a true/false result if the value matched the pattern or not - you'll have no information WHAT is missing in the value.
For example:
if(!value.Any(c => !char.IsLetter(c)){
throw new Exception("value must contain at least one non-letter")
}
I have to check whether the user entered string is in a particular format as below eg:
123-1234-1234567-1
ie after first 3 digit a hyphen, then after 4 digit another hyphen, after seven digit an hyphen then a single digit.
I used the below regular expression
#"^\(?([0-9]{3})\)?[-. ]?([0-9]{4})[-. ]?([0-9]{7})[-. ]?([0-9]{1})$"
It is working fine for above expression but it will also pass the expression without - also
eg:- 123-1234-1234567-1 //pass
123123412345671 //also getting pass.
The second string should fail. What change i should do in the regular expression to achieve the same?
You can simply use:
^\d{3}-\d{4}-\d{7}-\d$
If you want to allow dot and space also as delimiter then use:
^\d{3}[-. ]\d{4}[-. ]\d{7}[-. ]\d$
The problems is that you are having optional quantifier ? after [. ].
Remove them, and it should work fine
#"^\(?([0-9]{3})\)?[-. ]([0-9]{4})[-. ]([0-9]{7})[-. ]([0-9]{1})$"
Regex demo
The ? makes the preceding pattern optional as it matches 0 or 1 character. So in the second example the regex engine safely matches zero - to get to match the entire string
I am new to regular expressions and need a regular expression for address, in which user cannot enter repeating special characters such as: ..... or ,,,.../// etc and none of the special characters could be entered more than 5 times in the string.
...,,,....// =>No Match
Street no. 40. hello. =>Match
Thanks in advance!
I have tried this:
([a-zA-Z]+|[\s\,\.\/\-]+|[\d]+)|(\(([\da-zA-Z]|[^)^(]+){1,}\))
It selects all alphanumeric n some special character with no empty brackets.
You can use Negative lookahead construction that asserts what is invalid to match. Its format is (?! ... )
For your case you can try something like this:
This will not match the input string if it has 2 or more consecutive dots, commas or slashes (or any combination of them)
(?!.*[.,\/]{2}) ... rest of the regex
This will not match the input string if it has more than 5 characters 'A'.
(?!(.*A.*){5}) ... rest of the regex
This will match everything except your restrictions. Repplace last part (.*) with your regex.
^(?!.*[.,\/]{2})(?!(.*\..*){5})(?!(.*,.*){5})(?!(.*\/.*){5}).*$
Note: This regex may no be optimized. It may be faster if you use loop to iterate over string characters and count their occurences.
You can use this regex:
^(?![^,./-]*([,./-])\1)(?![^,./-]*([,./-])(?:[^,./-]*\2){4})[ \da-z,./-]+$
In C#:
foundMatch = Regex.IsMatch(yourString, #"^(?![^,./-]*([,./-])\1)(?![^,./-]*([,./-])(?:[^,./-]*\2){4})[ \da-z,./-]+$", RegexOptions.IgnoreCase);
Explanation
The ^ anchor asserts that we are at the beginning of the string
The negative lookahead (?![^,./-]*([,./-])\1) asserts that it is not possible to match any number of special chars, followed by one special char (captured to Group 1) followed by the same special char (the \1 backreference)
The negative lookahead (?![^,./-]*([,./-])(?:[^,./-]*\2){4}) ` asserts that it is not possible to match any number of special chars, followed by one special char (captured to Group 2), then any non-special char and that same char from Group 2, four times (five times total)
The $ anchor asserts that we are at the end of the string
A regular expression string to detect invalid strings is:
[^\w \-\r\n]{2}|(?:[\w \-]+[^\w \-\r\n]){5}
As C# string literal (regular and verbatim):
"[^\\w \\-\\r\\n]{2}|(?:[\\w \\-]+[^\\w \\-\\r\\n]){5}"
#"[^\w \-\r\n]{2}|(?:[\w \-]+[^\w \-\r\n]){5}"
It is much easier to find a string than to validate if a string does not contain ...
It can be checked with this expression if the string entered by the user is invalid because of a match of 2 special characters in sequence OR 5 special characters used in the string.
Explanation:
[^...] ... a negative character class definition which matches any character NOT being one of the characters listed within the square brackets.
\w ... a word character which is either a letter, a digit or an underscore.
The next character is simply a space character.
\- ... the hyphen character which must be escaped with a backslash within square brackets as otherwise the hyphen character would be interpreted as "FROM x TO z" (except when being the first or the last character within the square brackets).
\r ... carriage return
\n ... line-feed
Therefore [^\w \-\r\n] finds a character which is NOT a letter, NOT a digit, NOT an underscore, NOT a space, NOT a hyphen, NOT a carriage return and also NOT a line-feed.
{2} ... the preceding expression must match 2 such characters.
So with the expression [^\w \-\r\n]{2} it can be checked if the string contains 2 special characters in a sequence which makes the string invalid.
| ... OR
(?:...) ... none marking group needed here for applying the expression inside with the multiplier {5} at least 5 times.
[...] ... a positive character class definition which matches any character being one of the characters listed within the square brackets.
[\w \-]+ ... find a word character, or a space, or a hyphen 1 or more times.
[^\w \-\r\n] ... and next character being NOT a word character, space, hyphen, carriage return or line-feed.
Therefore (?:[\w \-]+[^\w \-\r\n]){5} finds a string with 5 "special" characters between "standard" characters.